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Math 11/L

ENGINEERING CALCULUS 1
MODULE 5: Application of Derivatives

Source: https://medium
...
EDWIN L
...


Overview

The topics in this module are
included in this section
...
It
helps you explore new ideas and
capabilities
...

It is a job aimed at evaluating
your mastery in acquiring learning
skills
...


Answer Key

This contains answers to all
activities in the module
...


MODULE 5 APPLICATION OF DERIVATIVES

1

ENGINEERING CALCULUS 1

OVERVIEW
COURSE DESCRIPTION:

[DATE]








CONTENTS:
The Differentials
The Differentials (Applications)
Parametric Equations
Indeterminate Forms
Maxima and Minima Problems
Related Rates Problems

An introductory course covering the core concepts of
limit, continuity and differentiability of functions involving one
or more variables
...


COURSE OUTCOME:
At the end of this course, you must be able to solve the problem that involve the application of
derivatives using the appropriate formula
...

INTRODUCTION
Good day future engineers! Welcome to Engineering Course
...
As you continue to do so in this
module, you can learn about the various ways in which the derivatives can be apply also in real life
situation
...
Just
like addition, subtraction, multiplication and division derivative is another concept in the field of
mathematics
...


LESSON PROPER
The differential of any function is equal to its derivative multiplied by the differential of the
independent variable
...

It follows from the above definition that all the fundamental formulas for derivatives become
differential formulas if we merely multiply through by 𝑑𝑥
...

EXAMPLE 5
...
1
Find the differential of function 𝑦 = 𝑥 3 − 2𝑥
SOLUTION:
𝑑𝑦 = 3𝑥 3−1 𝑑𝑥 − 2𝑥 1−1 𝑑𝑥
𝑑𝑦 = 3𝑥 2 𝑑𝑥 − 2𝑑𝑥
EXAMPLE 5
...
2
𝑧 2 −1

Find the differential of function 𝑦 = 𝑧 2 +1
SOLUTION:
𝑑𝑦 =

(𝑧 2 +1)2𝑧𝑑𝑧−(𝑧 2 −1)2𝑧𝑑𝑧
(𝑧 2 +1)2
4𝑧𝑑𝑧

=(𝑧 2 +1)2
EXAMPLE 5
...
3
Find the differential of function 𝑦 3 + 2𝑥𝑦 = 3
SOLUTION:
3𝑦 2 𝑑𝑦 + 2𝑥𝑑𝑦 + 2𝑦𝑑𝑥 = 0
−2𝑦𝑑𝑥

𝑑𝑦 = 3𝑦2+2𝑥

MODULE 5 APPLICATION OF DERIVATIVES

3

ENGINEERING CALCULUS 1

[DATE]

4
...
𝑦 = 3𝑥 4 − 15𝑥 3 + 𝑥 − 4
2
...
𝑢 = √2𝑦 − 𝑦 2
1

1

4
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
Compare your answer to the answer given, then try to answer
the questions again before you proceed to the next lesson
...


LESSON PROPER
Approximate Formulas
Very often we wish to compute, or to estimate within safe limits, the change in the value of a
function caused by a small change in the value of the independent variable
...

In any approximate computation, the amount by which the computed value of the function differs
from the true is called error of the computation
...

Example 5
...
1
Find an approximate formula for the area of a narrow circular ring
...
2
...
73
SOLUTION
Put 𝑦 = √𝑥, from which
𝑑𝑥

𝑑𝑦 = 2

√𝑥

For 𝑥 we choose which is close to 8
...
Choose
𝑥=9

𝑎𝑛𝑑

∆𝑥 = 𝑑𝑥 = −0
...
73
...
27
2√9

=

−0
...
045

√8
...
045 = 𝟐
...
To five decimal places the correct value is 2
...
2
...
, using
1
boards 2 𝑖𝑛
...

SOLUTION
The volume of cube of a length 𝑥 is,
𝑉 = 𝑥 3 , from which
𝑑𝑉 = 3𝑥 3 𝑑𝑥 = 3𝑥 3 ∆𝑥
Assume that 𝑑𝑥 = ∆𝑥 =

0
...


0
...
) ( 12 𝑓𝑡
...
2
...
with a maximum error of 0
...
Find
the approximate maximum error in the computed volume
...
1 𝑖𝑛
...
2 (0
...
𝟑
Propagated and Relative Error
If a measured value x is used to compute another value 𝑓(𝑥), the difference of 𝑓(𝑥 + ∆𝑥) and
𝑓(𝑥) or ∆𝑦 is the propagated error
...

If the propagated error is given in relative terms if we compare 𝑑𝑉 with 𝑉
...


𝑑𝑉
𝑉

is called the

Example 5
...
5
The radius of the ball bearing is measured to be 0
...
If the measurement is correct within
0
...

SOLUTION
a)

4

𝑉 = 3 𝜋𝑟 3 → 𝑑𝑉 = 4𝜋𝑟 2 𝑑𝑟

where 𝑟 = 0
...
02

𝑑𝑉 = 4𝜋(0
...
02)
= ±𝟎
...
02)
0
...
𝟓%

MODULE 5 APPLICATION OF DERIVATIVES

6

ENGINEERING CALCULUS 1

[DATE]

Example 5
...
6
A right circular cylinder has a height of 24 ft
...
If the error measuring is 3 inches,
a) what is the possible error in volume b) what is the relative error?
SOLUTION
a)
...
25𝑓𝑡) + 𝜋𝑟 2 (±0
...
25𝑓𝑡) + 𝜋(82 𝑓𝑡 2 )(±0
...
𝟖 𝒇𝒕𝟑
b)
...
5𝑓𝑡 3
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 =
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 =

𝑑𝑉
𝑉
±351
...
5 𝑓𝑡 3

= ±𝟎
...
2 Practice Problem
Answer the following problems:
1
...

2
...

3
...

4
...
01 𝑖𝑛
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
Compare your answer to the answer given, then try to answer
the questions again before you proceed to the next lesson
...


LESSON PROPER
3
...
These equation is called parametric equation of a
curve
...
When dealing parametric
equations, it is convenient to use differentials in finding derivatives, particularly the derivatives of higher
order
...
3
...
3
...
3a Practice Problem
Find the first and second derivatives of 𝑦 with respect to 𝑥 from the parametric equations given
...
𝑥 = 1 + 𝑡 2 , 𝑦 = 4𝑡 − 3
2
...
𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 2 + 1
4
...
Please answer it all first before checking, this is important for
self-assessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice
for a few times
...
your
Take
a breather
beforegiven,
you proceed
the
If you got 0 –You
1, it is
okay,finished
do not worry
...


lesson 4
...
2 Parametric Equation (Application)
3
...
1 Rectilinear Motion
Another application of finding derivatives is to compute velocities and accelerations of moving
objects
...
That is, the
acceleration is the derivative of the velocity with respect to time 𝑡
...
So we
have
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑎 =

𝑑𝑣
𝑑𝑡

=

𝑑2 𝑥
𝑑𝑡 2

𝑑𝑣

= 𝑣 𝑑𝑥

Example 5
...
3
The position of an object at time t is given by 𝑥 = 3𝑡 3 − 6𝑡 2 + 4𝑡 − 2, where 𝑥 is in feet and 𝑡 is
in seconds
...
2
...
An important instance
of uniformly accelerated motion arise when a body moves near the earth’s surface in a vertical
straight line
...

𝑎=

𝑑𝑣
𝑑𝑡

=𝑔;

1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡;

𝑣 2 = 𝑣𝑜 2 + 2𝑔𝑥

Example 5
...
4
A ball is dropped from the balloon at a height of 640𝑓𝑡
...

SOLUTION
With 𝑔 = 32
1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡
1

𝑥 = 2 (−32)𝑡 2 + 96𝑡, from which
𝑑𝑥 = (−32𝑡 + 96)𝑑𝑡 =

𝑑𝑥
𝑑𝑡

= 𝑣 = −32𝑡 + 96

At the highest point, 𝑣 = 0; hence 𝑡 = 3,
1
𝑥 = (−32)(3)2 + 96(3) = 144𝑓𝑡
2

Thus the height above the starting point is 144𝑓𝑡
...
:
1

𝑥 = 2 𝑔𝑡 2 + 𝑣𝑜 𝑡
640 = 16𝑡 2 − 96𝑡
16(𝑡 2 − 6𝑡 − 40) = 0; 𝑡 = −4
Thus the ball is in the air for 𝟏𝟎 𝒔𝒆𝒄
...
2
...

With the starting point as origin and the y-axis is positive upward, the initial condition are;
𝑥 = 0,
𝑦 = 0,
𝑣𝑥 = 𝑣𝑜 cos 𝛼,
𝑣𝑦 = 𝑣𝑜 sin 𝛼 when 𝑡 = 0
The force of gravity acts vertically downward; there is no horizontal force
...
3
...
How far does an arrow flies?
SOLUTION:
When the arrow hit the ground at maximum distance
𝑦 = 0
...
8 𝑠𝑒𝑐
𝑥 = 𝑣𝑜 𝑡 cos 𝛼
𝑥 = 200(8
...
𝟓𝒇𝒕
3
...
4 Velocity and Acceleration in Curvilinear Motion
If a point moves in a plane curve, its coordinates are functions of time:
𝑥 = 𝑓(𝑡),
𝑦 = 𝑔(𝑡)
Thus the total velocity is the vector sum of the velocities parallel to the axis
𝑑𝑥

𝑑𝑦

𝑑𝑡

𝑑𝑡

𝑣 = √𝑣𝑥 2 + 𝑣𝑦 2 = √( )2 + ( )2
inclined to the x-axis at an angle 𝛼 such that
𝑣𝑦

tan 𝛼 = 𝑣

𝑥

The acceleration is the vector 𝛼 whose components, parallel to the axis are
𝑑𝑦 2

𝑑2 𝑥

𝑎 = √𝑎𝑥 2 + 𝑎𝑦 2 = √ 𝑑𝑡 2 + 𝑑𝑡 2

inclined to the x-axis at an angle 𝛽 such that
𝑎𝑦

tan 𝛽 = 𝑎

𝑥

Example 5
...
6
Given the parametric equation below determine the total velocity and acceleration at 𝑡 = 2
𝑥 (𝑡) = 𝑡 3 − 3𝑡 2 + 𝑡

𝑦(𝑡) = 𝑡 2 − 2𝑡 + 3

and

SOULUTION:
For velocity
𝑥 (𝑡) = 𝑡 3 − 3𝑡 2 + 𝑡 from which

𝑑𝑥

𝑦(𝑡) = 𝑡 2 − 2𝑡 + 3 from which

𝑑𝑦

𝑣 = √(

𝑑𝑡

𝑑𝑡

= 3𝑡 2 − 6𝑡 + 1
= 2𝑡 − 2

𝑑𝑥 2
𝑑𝑦
) + ( )2 = √(3(2)2 − 6(2) + 1)2 + (2(2) − 2)2 = √5
𝑑𝑡
𝑑𝑡

MODULE 5 APPLICATION OF DERIVATIVES

11

ENGINEERING CALCULUS 1

[DATE]

For Acceleration
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑡

= 3𝑡 2 − 6𝑡 + 1 from which
= 2𝑡 − 2 from which

𝑑2 𝑥
𝑑𝑡 2

𝑑2 𝑥
𝑑𝑡 2

= 6𝑡 − 6

=2

2
𝑑 2 𝑥 𝑑𝑦
𝑎 = √ 2 + 2 = √(6(2) − 6)2 + 22 = 𝟒√𝟐
𝑑𝑡
𝑑𝑡

5
...
The position of an object at time t is given by 𝑥 = 34𝑡 2 − 8𝑡 + 2, where 𝑥 is in feet and 𝑡 is in
seconds
...
A ball is thrown upward and rises 9 𝑓𝑡
...
Find the total time taken for the ball to
return to the starting point
...
A point describe parabola 𝑦 2 = 4𝑥 + 1, with a constant velocity, 𝑣𝑦 = 4
...

Check your answer in the answer key
...

If you got 3 items correct, very good, you already understand the concept
...

If you got 0 – 1, it is okay, do not worry
...


MODULE 5 APPLICATION OF DERIVATIVES

12

ENGINEERING CALCULUS 1

[DATE]

Lesson

4

Indeterminate Forms

In this lesson you will learn how to evaluate the limits of indeterminate forms equations using
derivatives
...
If 𝑓 (𝑎) = 𝑓(𝑏), then there
must be at least one number 𝑐 in (𝑎, 𝑏) such that 𝑓 ′(𝑐 ) = 0
...
1 Example diagram for Rolle’s Theorem
In the first two diagrams, there is only one possible value of 𝑐 such that 𝑓 ′(𝑐 ) = 0
...
The fourth diagram shows a constant function, so its derivative is always 0
...
Now, let’s look at some pictures where Rolle’s
Theorem does not apply:

Figure 5
...
2, in all three cases the derivative is never been zero
...
1
Find the point where the derivative is equal to zero of the equation 𝑓 (𝑥 ) = 𝑥 2 − 2𝑥 + 1 and
between points (−1,3)
...
Then there’s at least one
number 𝑐 in (𝑎, 𝑏) such that
𝑓 ′ (𝑐 ) =

𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎

Figure 5
...
In the above picture, there are actually
two tangents that work—the x-coordinates are at 𝑐𝑜 and 𝑐1
...

Example 4
...
Use Mean Value Theorem
...

Example 4
...
The product is then said to take the
indeterminate form 0⋅ ∞
...

0

Example 4
...

Example 4
...
The transformation required are simple:
1
sin3 𝑥
1 − sin3 𝑥
lim+ (sec3 𝑥 − tan3 𝑥) = lim+ ( 3 −
)
=
lim
𝜋
𝜋
𝜋+ cos 3 𝑥
cos 𝑥 cos 3 𝑥
𝑥→
𝑥→
𝑥→
2

2

2

By using theorem 23
lim+

𝜋
𝑥→
2

−3 sin2 𝑥 cos 𝑥
= lim tan 𝑥 = −∞
−3 cos 2 𝑥 sin 𝑥 𝑥→𝜋+
2

The Indeterminate Form 00 , ∞0 , 1∞
Consider the function 𝑦 = 𝑓(𝑥)𝐹(𝑥) , the function is said, in the respective cases, assume the
indeterminate form 00 , ∞0 , 1∞
...

Example 4
...
Then
ln 𝑦 =

ln(1−cos 𝑥)
ln 𝑥

Therefore,
lim+ ln 𝑦 = lim+

𝑥→0

𝑥→0

ln(1−cos 𝑥)
ln 𝑥
sin 𝑥

𝑥
= lim+ 1−cos
1

𝑥→0

𝑥

𝑥 sin 𝑥

= lim+ 1−cos 𝑥
𝑥→0

= lim+
𝑥→0

sin 𝑥+𝑥 cos 𝑥
sin 𝑥

= 1+1 = 2
From 𝑙𝑛𝑦 → 2 it follows that 𝑦 = 𝑒 2
...
4 Practice Problem
Evaluate the limit of the following:
1
...
lim sin−1 𝛽
𝛽→0

1

1

3
...
lim (cot 𝑥 − 𝑒𝑥 −1)
𝑥→0

Check your answer in the answer key
...

If you got 4 - items correct, very well, you already understand the concept
...

If you got 0 – 1, it is okay, do not worry
...


MODULE 5 APPLICATION OF DERIVATIVES

17

ENGINEERING CALCULUS 1

[DATE]

Lesson

Maxima and Minima Problems

5

In this lesson you will learn how to evaluate the derivatives of a Hyperbolic functions
...
1 Maxima and Minima
Max-Min Theorem: If 𝑓 is continuous on (𝑎, 𝑏), then 𝑓 has at least one maximum and one minimum
on (𝑎, 𝑏)
...


In the first graph, the function attains its maximum at 𝑥 = 𝑐 and its minimum at 𝑥 = 𝑑
...
The third
graph has a maximum at 𝑥 = 𝑏 but the minimum is at both 𝑥 = 𝑐 and 𝑥 = 𝑑
...
Finally, the fourth graph shows a
constant function, which is continuous; in fact, every point in the interval (𝑎, 𝑏) is both a maximum and
a minimum, since the function never goes above or below the constant value 𝑐
...
5
...
Now as 𝑥
(increasing) passes through −1, 𝑦′ changes from positive to negative; thus 𝑦 assumes the maximum
19
value
...


MODULE 5 APPLICATION OF DERIVATIVES

18

ENGINEERING CALCULUS 1

[DATE]

5
...
5
...
This result finds
application in a great variety of problems, some of which will now considered
...
In practice, the value that gives the desired maximum or minimum can be selected at
once by inspection
...
5
...
by cutting equal squares out of the
corner and turning up the sides
...

SOLUTION
Let 𝑥 be the length of the side of each of the squares cut out
...

𝑉𝑚𝑎𝑥 = 2(10 − 4)(16 − 4) = 𝟏𝟒𝟒𝒊𝒏
...
5
...

SOLUTION
The area of the rectangle is
𝐴 = 4𝑥𝑦
Where x and y are connected by the relation
𝑥 2 + 𝑦 2 = 𝑎2
By substituting 𝑦 = √𝑎2 − 𝑥 2 , we find
𝐴 = 4𝑥√𝑎2 − 𝑥 2
So that
4𝑥 2

𝐴′ = 4√𝑎2 − 𝑥 2 − √𝑎2

−𝑥 2

=

4𝑎2 −8𝑥 2
√𝑎2 −𝑥 2
1

Setting 𝐴′ = 0 we get 4𝑎2 − 8𝑥 2 = 0, 𝑥 = 2 √2𝑎
𝑨𝒎𝒂𝒙 = 𝟐𝒂𝟐

MODULE 5 APPLICATION OF DERIVATIVES

19

ENGINEERING CALCULUS 1

[DATE]

Example 5
...
4
Find the area of a largest rectangular frame is to be made from a piece of wire 120 𝑐𝑚 long
...
5 Practice Problem
Answer the following problem:
1
...
What positive number added to its reciprocal gives the minimum sum?
3
...
Find the minimum value of the sum of their cubes
...
A box is to be made of a piece of card board 9 𝑖𝑛
...
Find the volume of the largest box that can be made this way
...
Please answer it all first before checking, this is important for selfassessment
...

If you got 2 – 3 items correct, good, you understand most of the concepts, you only need to practice for a
few times
...
Compare your answer to the answer given, then try to answer the
questions again before you proceed to the next lesson
...


LESSON PROPER
6
...


Example 5
...
1
A balloon, leaving the ground 60𝑓𝑡
...

How fast is the balloon receding from the observer, after 8 𝑠𝑒𝑐
...
6
...
If the bridge is 30𝑓𝑡
...

later?
SOLUTION
In 𝑡 𝑠𝑒𝑐
...
By elementary geometry, the
distance between them is
𝑠 = √(5𝑡)2 + (10𝑡) + (30)2
= √125𝑡 2 + 900;
𝑑𝑠
𝑑𝑡

=

125𝑡
√125𝑡+900

After 3 𝑠𝑒𝑐
...
6
...
At what rate is the boat approaching the wharf when there is 25𝑓𝑡 of rope out?
SOLUTION
Let 𝑥 denote the distance of the boat from the wharf, 𝑟 the length of rope
...
To do this, as suggested above, we express 𝑥 in terms of 𝑟 (implicitly or explicitly) and
differentiate with respect to 𝑡:
𝑑𝑥

𝑥 = √𝑟 2 − 400,
Substitute 𝑟 = 25,
𝑑𝑥
𝑑𝑡

𝑑𝑟

𝑟

𝑑𝑟
𝑑𝑡

√𝑟 2 −400

= −4

𝑑𝑡

=

𝑑𝑡

=

−100

=

√225

𝟐𝟎
𝟑

𝒇𝒕/𝒔𝒆𝒄

Example 5
...
4
Water is flowing into a conical reservoir 20𝑓𝑡 deep and 10𝑓𝑡 across the top, at rate of
15𝑓𝑡/𝑚𝑖𝑛
...
deep
...
𝟏𝟗𝒇𝒕/𝒎

MODULE 5 APPLICATION OF DERIVATIVES

22

ENGINEERING CALCULUS 1

[DATE]

6
...
6
...
So the equation in the box above
becomes 𝑃(𝑡) = 1000𝑒 𝑘𝑡
...
We do know that 𝑃 = 64000
when 𝑡 = 3, so let’s plug this in:
64000 = 1000𝑒 3𝑘
𝑘 = 2 ln 2
This means
𝑃 (4) = 1000𝑒 2𝑙𝑛2(4) = 𝟐𝟓𝟔, 𝟎𝟎𝟎 𝒓𝒂𝒃𝒃𝒊𝒕𝒔
400,000 = 1000𝑒 2𝑙𝑛2(𝑡)
𝑙𝑛400
𝑡 = 2𝑙𝑛2
𝒕 = 𝟒
...
That’s the power of exponential
growth
...
6 Practice Problem
Answer the following problem:
1
...
If the radius of the tank is
4𝑓𝑡
...
Water flows into a vertical tank at 12𝑓𝑡 3 /𝑚𝑖𝑛 ;the surface rises in 6 𝑖𝑛/𝑚𝑖𝑛
...

3
...
How fast does the
end of his shadow move?
4
...
Find how fast the surface rises, if the water flows in
at the rate of 12𝑓𝑡 3 /𝑚𝑖𝑛

Check your answer in the answer key
...

If you got 4 items correct, very well, you already understand the concept
...

If you got 0 – 1, it is okay, do not worry
...


MODULE 5 APPLICATION OF DERIVATIVES

24

ENGINEERING CALCULUS 1

[DATE]

Assessment
This part is a graded assessment, I will assess if you have learned the whole module
...
Write your answer in a long bond paper together with front page (must include
course code and the title, title of the lesson, title of activity, name of student and your block, date of
submission, name of your professor)
...
5 inches’ border
including the front page
...
Once you’re done answering, send it to me
via Google Classroom in a PDF file
...
Good Luck
Problem Set No
...
𝑧 = √4 − 3𝑥
1−𝑠

2
...

1

3
...
𝑥 = √1 − 𝑡; 𝑦 = 𝑡 3 − 3𝑡
5
...
lim
1

1−sin 𝑥

𝑥→ 𝜋 cos 5𝑥
2

7
...
lim ( sin2 𝑥 −
𝑥→0

sin 𝑥
𝑥3

)

9
...
,the hypotenuse is 5𝑓𝑡
...
Find
the approximate change in altitude when hypotenuse is changed by small amount ∆ℎ
...
The diameter of circle is measured and found to be 6𝑓𝑡 with a maximum error of 0
...
Find the
approximate maximum error in the computed area
...
Find the change in the lateral surface area of a right circular cone, with radius of base fixed as 𝑟,
when the altitude ℎ changes by a small amount ∆ℎ
...
The position of an object at time t is given by 𝑥 = √3𝑡 − 4 +

1
√𝑥

+ 4, where 𝑥 is in feet and 𝑡 is in

seconds
...
A ball is thrown upward from the ground with a speed of 40𝑓𝑡/𝑠𝑒𝑐; at the same instant another
ball is dropped (from rest) from a height of 100𝑓𝑡
...

14
...
Find two numbers whose sum is 𝑎, if the product of one by the cube of the other is to be maximum
...
A sphere is cut to shape a circular cone
...
Find the most economical proportions for a cylindrical cup
...
A rectangular field of fixed area is to be enclosed and divided into three lots by parallel to one sides
...
A wall 10 𝑓𝑡 high is 8𝑓𝑡 from the house
...

20
...
If water flows in at the rate of
3𝑓𝑡
...
deep
...
A ladder 20 𝑓𝑡 long leans against the wall
...

22
...
How fast does the
shadow lengthen?
23
...
Another train, starting from the same point at
2 𝑃𝑀, travels east at 50𝑚𝑖/ℎ𝑟, how fast the two trains are separating at 3𝑃𝑀
...
A light at eye level stands 20𝑓𝑡 from the house and 15 𝑓𝑡 from the path leading from the house to
the street
...
How fast does his shadow move along the wall
when he is 5𝑓𝑡 from the house?

MODULE 5 APPLICATION OF DERIVATIVES

26

ENGINEERING CALCULUS 1

[DATE]

Answer Key
Answers for 5
...
𝑑𝑦 = 12𝑥 3 𝑑𝑥 − 15𝑥 2 𝑑𝑥 + 𝑑𝑥
3

2
...
𝑑𝑢 = (1 − 𝑦)(2𝑦 − 𝑦 2 )−2 𝑑𝑦
1

4
...
2 Practice Problem:
1
...
𝑐𝑖𝑟𝑐𝑢𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡 𝑥 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
3
...
04
4
...
52𝑖𝑛2
Answer for 4
...

2
...

4
...
3b Practice Problem:
1
...
1
...
𝑣𝑥 = 4;𝑎𝑥 = 2
Answer for 5
...
1/2
2
...
0
4
...
5 Practice Problem:
1
...
1
1

3
...
54𝑖𝑛
...
6 Practice Problem:
1
...
18 𝑓𝑡/𝑚𝑖𝑛
2
...
76 𝑓𝑡
3
...
0
...
Louis Leithold , The calculus 7 7th edition
2
...
Love and Earl D
...
Dit Gallesania, Engineering mathematics volume 2, Third Edition

MODULE 5 APPLICATION OF DERIVATIVES

28



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